Home | | **Dynamics of Machines** | Inertia Forces in a Reciprocating Engine, Considering the Weight of connecting Rod

In a reciprocating engine, let OC be the crank and PC, the connecting rod whose centre of gravity lies at G. The inertia forces in a reciprocating engine may be obtained graphically as discussed below:

**INERTIA FORCES IN A RECIPROCATING
ENGINE, CONSIDERING THE WEIGHT OF CONNECTING ROD**

In a
reciprocating engine, let *OC* be the
crank and *PC*, the connecting rod
whose centre of gravity lies at *G*.
The inertia forces in a reciprocating engine may be obtained graphically as
discussed below:

**1. **First
of all, draw the acceleration diagram** ***OCQN***
**by Klien’s construction. We** **know
that the acceleration of the piston *P*
with respect to *O*,

acting in the direction from *N* to *O*.
Therefore, the inertia force *F*_{I}
of the reciprocating parts will act in the opposite direction as shown in Fig.
15.22.

**Fig. 15.22. **Inertia forces is reciprocating
engine, considering the weight of connecting** **rod.

2. Replace
the connecting rod by dynamically equivalent system of two masses as discussed
in Art. 15.12. Let one of the masses be arbitrarily placed at *P*. To obtain the position of the other
mass, draw *GZ* perpendicular to *CP* such that *GZ* = *k*, the radius of
gyration of the connecting rod. Join *PZ*
and from *Z* draw perpendicular to *DZ* which intersects *CP* at *D*. Now, *D* is the position of the second mass.

**Note: **The position of the second mass may
also be obtained from the equation,

*GP
*×*
GD* = *k*^{2}

**3. **Locate the points** ***G*** **and** ***D*** **on** ***NC*** **which is the
acceleration image of the** **connecting
rod. This is done by drawing parallel lines from *G* and *D* to the line of
stroke *PO*. Let these parallel lines
intersect* NC *at* g *and* d *respectively.
Join* gO *and* dO*. Therefore,* *acceleration
of *G* with respect to *O*, in the direction from *g* to *O*,

aGO = aG = ω^{2} × gO

and acceleration of *D* with respect to *O*, in the direction from *d*
to *O*,

aDO =
aD = ω^{2} × dO

From *D*, draw *DE* parallel to *dO* which intersects the line of stroke *PO* at *E*. Since the accelerating forces on the masses at *P* and *D* intersect at *E*, therefore
their resultant must also pass through *E*.
But their resultant is equal to the accelerang force on the rod, so that the
line of action of the accelerating force on the rod, is given by a line drawn
through *E* and parallel to *gO*, in the direc- tion from *g* to *O.*
The inertia force of the connecting rod *F*_{C}* *therefore acts through* E *and in the opposite direction as
shown in Fig. 15.22. The* *inertia
force of the connecting rod is given by

A little consideration will show that the forces acting on the connecting rod are :

**( a)** Inertia force of the reciprocating
parts (

**( b)** The side thrust between the
crosshead and the guide bars (

right angles to line of stroke *PO,*

**( c)** The weight of the
connecting rod

(*W*
_{C} = *m*_{C}.*g*), **( d) **Inertia force of the connecting
rod (

**( e) **The radial force (

Now, produce the lines of action of *F*_{R} and *F*_{N} to intersect at a point *I*, known as instantaneous centre. From *I* draw *I X* and *I Y* , perpendicular to the lines of
action of *F*_{C} and *W* _{C}. Taking moments about *I*, we have

*F*_{T}* *×* IC *=* F*_{I}* *×* IP *+* F*_{C}* *×* I X *+* W *_{C}* *×* I Y *...**( ii)**

The value of *F*_{T} may be obtained from this equation and from the force
polygon as

shown in Fig. 15.22, the forces *F*_{N} and *F*_{R} may be calculated. We know that, torque exerted on
the crankshaft to overcome the inertia of the moving parts = *F*_{T} × *OC*

1 **Analytical Method for Inertia Torque**

The effect of the inertia of the
connecting rod on the crankshaft torque may be obtained as discussed in the
following steps:

**Fig. 15.23. **Analytical method for inertia
torque.

**1.**The mass of the connecting rod (*m* _{C}) is divided into two masses. One of the mass is
placed at the crosshead pin *P* and the
other at the crankpin *C* as shown in
Fig. 15.23, so that the centre of gravity of these two masses coincides with
the centre of gravity of the rod

*G*.

**2. **Since the inertia force due to the
mass at *C* acts radially outwards
along the crank *OC*, therefore the
mass at* C *has no effect on the
crankshaft torque.

**3. **The inertia force of the mass at *P* may be obtained as

follows: Let
connecting rod, mC = Mass of the

*l* =
Length of the connecting rod,

*l*_{1} =
Length of the centre of gravity of the connecting rod from P.

**4. **In
deriving the equation** ( ii) **of the torque exerted on the
crankshaft, it is

The correcting torque *T'* may be applied to the system by two
equal and opposite forces *F*Yacting
through *P* and *C*. Therefore,

**Q.***The
crank and connecting rod lengths of an engine are 125 mm and 500 mm***
***respectively. The mass of the connecting rod is 60 kg and
its centre of gravity is 275 mm from the crosshead pin centre, the radius of
gyration about centre of gravity being 150 mm.*

*If the engine speed is 600 r.p.m. for a crank position of
45° from the inner dead centre, determine, using Klien’s or any other
construction 1. the acceleration of
the piston; 2. the magni- tude,
position and direction of inertia force due to the mass of the connecting rod.*

**Solution. **Given :** ***r*** **=** ***OC*** **= 125 mm ;** ***l*** **=**
***PC*** **= 500 mm;** ***m***
**_{C}** **= 60 kg ;** ***PG*** **= 275

mm ;

*m *_{C}* *= 60 kg ;* PG *= 275
mm ;* k*_{G}* *= 150 mm ;* N *= 600 r.p.m. or* ** *= 2*
*× 600/60 = 62.84 rad/s ;* * = 45°

**1. Acceleration of the piston**

Let *a*_{P}* *= Acceleration of the piston.

First of all, draw the configuration
diagram *OCP*, as shown in Fig. 15.24,
to some suitable scale, such that

*OC
*=*
r *= 125 mm ;* PC *=* l *= 500 mm ; and* **
*= 45°.

Now, draw the Klien’s acceleration
diagram *OCQN,* as shown in Fig. 15.24,
in the same manner as already discussed. By measurement,

*NO
*= 90 mm = 0.09 m

Acceleration of the piston*,*

*a*_{P}* *=* *ω^{2}* *×* NO *= (62.84)^{2}
× 0.09 = 355.4 m/s **Ans.**

**2. The magnitude,
position and direction of inertia force due to the mass of the connecting rod**

The magnitude, postition and
direction of the inertia force may be obtained as follows:

**( i)
**Replace the connecting rod by
dynamical equivalent system of two masses,

**( ii)
**Locate the points

*gO
*= 103 mm = 0.103 m

Acceleration of *G*, *a*_{G}* *=*
*ω^{2}* *×* gO*, acting in
the direction from* g *to* O*.

**( iii)
**From point

*F*_{C}* *=* m*_{C}* *×*
*ω^{2}* *×* gO *= 60 ×
(62.84)^{2}* *× 0.103 = 24 400
N = 24.4* *kN **Ans. ( iv)** From point

inertia force of

the connecting rod and acts in the
opposite direction of *gO*.

**Q. ***The following data refer to a steam
engine:*

*Diameter of piston = 240 mm; stroke *=*
600 mm *;* length of connecting rod =
1.5 m ; mass of reciprocating parts = 300 kg; mass of connecting rod = 250 kg;
speed = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 500 mm
; radius of gyration of the connecting rod about an axis through the centre of
gravity = 650 mm.*

*Determine the magnitude and direction of the torque exerted
on the crankshaft when the crank has turned through 30° from inner dead centre.*

**Solution. **Given :** ***D*** **= 240 mm = 0.24 m ;** ***L***
**= 600 mm or** ***r*** **=** ***L*/2 = 300 mm =** **0.3 m ; *l* = 1.5 m ; *m*_{R}
= 300 kg ; *m*_{C} = 250 kg ; *N* = 125 r.p.m. or ω = 2π × 25/60 = 13.1 rad/s ; *GC* = 500 mm = 0.5 m ; *k*_{G}
= 650 mm = 0.65 m ; θ = 30°

The
inertia torque on the crankshaft may be determined by graphical method or
analytical method as discussed below:

**1. Graphical method**

First of
all, draw the configuration diagram *OCP*,
as shown in Fig. 15.25, to some suitable scale, such that

*OC
*=*
r *= 300 mm ;* PC *=* l *= 1.5 m ; and angle* POC *=* *θ* *= 30°.

Now draw the Klien’s acceleration
diagram *OCQN*, as shown in Fig. 15.25,
and complete the figure in the similar manner as discussed in Art. 15.14.

By measurement; *NO* = 0.28 m ; *gO* = 0.28 m
; *IP* = 1.03 m ; *I X* = 0.38 m ; *I Y* = 0.98
m, and *IC* = 1.7 m.

We know that inertia force of
reciprocating parts,

**Q. ***The connecting rod of an internal
combustion engine is 225 mm long and has a*** ***mass 1.6 kg. The mass of the piston
and gudgeon pin is 2.4 kg and the stroke is 150 mm. The cylinder bore is 112.5
mm. The centre of gravity of the connection rod is 150 mm from the small end.
Its radius of gyration about the centre of gravity for oscillations in the
plane of swing of the connect- ing rod is 87.5 mm. Determine the magnitude and
direction of the resultant force on the crank pin when the crank is at 40° and
the piston is moving away **from
inner dead centre under an effective gas presure of 1.8 MN/m ^{2}. The
engine speed is 1200 r.p.m.*

**Solution. **Given :** ***l*** **=** ***PC*** **= 225 mm = 0.225 m;** ***m*_{C}** **= 1.6 kg;** ***m*** **_{R}** **= 2.4 kg;** ***L*** **=**
**150 mm or *r* = *L*/2 = 75 mm = 0.075 m ; *D* = 112.5 mm = 0.1125 m ; *PG* = 150 mm ; *k*_{G} = 87.5 mm = 0.0875 m ; θ = 40° ; *p* = 1.8 MN/m^{2} = 1.8 × 10 ^{6} N/m^{2} ; *N* = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s

First of all, draw the configuration
diagram *OCP*, as shown in Fig. 15.27
to some suitable scale, such that *OC*
= *r* = 75 mm ; *PC* = *l* = 225 mm ; and θ = 40°.

Now, draw the Klien’s acceleration
diagram *OCQN*. Complete the diagram in
the same manner as discussed earlier. By measurement,

*NO *=
0.0625 m ;* gO *= 0.0685 m ;* IC *= 0.29 m ;* IP *= 0.24 m ;* I Y *=
0.148 m ; and* IX *=* *0.08 m

We know that force due to gas
pressure,

Let
us now find the values of *F*_{N}
and *F*_{R} in magnitude and
direction. Draw the force polygon as shown in Fig. 15.25.

By measurement, *F*_{N} = 3550 N; and *F*_{R}
= 7550 N

The
magnitude and direction of the resultant force on the crank pin is given by *F*_{Q} , which is the resultant
of *F*_{R} and *F*_{T}.

By measurement, *F*_{Q} = 13 750 N **Ans.**

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